# What is the derivative of f(t) = (e^(t^2-1)+3t, -t^3+t ) ?

Jan 10, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 {t}^{2} + 1}{2 t {e}^{{t}^{2} - 1} + 3}$

#### Explanation:

$x \left(t\right) = {e}^{{t}^{2} - 1} + 3 t$
$y \left(t\right) = - {t}^{3} + t$

$x ' \left(t\right) = 2 t {e}^{{t}^{2} - 1} + 3$
$y ' \left(t\right) = - 3 {t}^{2} + 1$

The derivative of a parametric function can be found through

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}} = \frac{y ' \left(t\right)}{x ' \left(t\right)} = \frac{- 3 {t}^{2} + 1}{2 t {e}^{{t}^{2} - 1} + 3}$